Invariants of diffeomorphisms of four-manifolds

More gauge theory coming at you.  This post is based on Ruberman's paper "An obstruction to smooth isotopy in dimension 4", and inspired by a conversation with John Baldwin.

I want to talk about an idea in gauge theory that is slightly less advertised, not too complicated to describe, and still really cool.  Let's say we have a closed four-manifold $X$ (simply-connected, smooth, oriented, connected, large $b^+_2$, all the good things).  A natural question is how to distinguish self-diffeomorphisms of $X$ up to isotopy?  Freedman and Quinn proved that if two diffeomorphisms of $X$ induce the same map on homology, then they are isotopic through homeomorphisms.  Totally cool, but they may not be isotopic through diffeomorphisms!  It turns out we can use gauge theory to make this distinction.  Get pumped!  

First, here's the basic idea.  If we fix a metric $g$ on $X$ and an $SO(3)$-bundle $E$ with non-trivial $w_2$, then recall that we get a moduli space $\mathcal{M}(X,E, g)$ of ASD connections.  (It's not so important to us what this means, but you can see here for more details.)  For generic metrics, $\mathcal{M}(X,E,g)$ is a nice smooth orientable manifold, and if it's 0-dimensional, we can define the Donaldson invariant to just be the count of ponits.  For simplicity, we'll omit the bundle from the notation.

If $f :X \to X$ is a diffeomorphism, then we could try to compare the Donaldson invariants associated to $g$ and to $f^*g$.  But, they should be the same, since the counts are supposed to be topological invariants, so we need a slightly different trick - studying a path of metrics between $g$ and $f^*g$.     

Silly observation: If the dimension of $\mathcal{M}(X,g)$ is negative, then the moduli space is empty, since it is negative-dimensional.  

Interesting observation: Suppose for simplicity that the dimension of $\mathcal{M}(X,g)$ is $-1$.  If we consider a generic *path* of metrics $g_0$ from $g_0$ to $g_1$ and look at $\bigcup_{0 \leq t \leq 1} \mathcal{M}(X,g_t)$, then this is roughly a 1-dimensional family of $-1$-dimensional manifolds, and $1 + -1 = 0$.  But, seriously, we get a 0-dimensional moduli space $\mathcal{M}(X,\{g_t\})$ and we can take a signed count of the number of points per usual.  If you've heard the term "family" gauge theory, this is what it refers to: basically looking at a parameterized family of moduli spaces.

Good news: If we pick $g_0, g_1$, then $ \# \mathcal{M}(X,\{g_t\})$ is independent of the generic path between them.  (The key idea is that the space of metrics is contractible, so any two paths cobound a disk in the space of metrics.  There's a moduli space parameterized by metrics in that disk which is a 1-manifold with boundary - the boundary is the difference of $\# \mathcal{M}(X,\{g_t\})$ and $\# \mathcal{M}(X,\{g'_t\})$ for the two different paths $g_t$ and $g'_t$.  The signed count of points coming from the boundary of a 1-manifold is zero, so the two counts must agree.)  

Better news:  Let $f : X \to X$ be a diffeomorphism.  Fix a metric $g$ and a path $g_t$ from $g$ to $f^*g$.   We can define an invariant of the diffeomorphism $f$ by the count $\# \mathcal{M}(X,\{g_t\})$.  This in principle could depend on the metric $g$ we start with, but actually it doesn't!  This is explained in the next paragraph if you care.    

Ignore this paragraph if you're not interested in the proof of invariance:  Here's the idea.  Fix metrics $g$ and $k$.  Let $g_t$ and $k_t$ be paths from $g$ to $f^*g$ and from $k$ to $f^*k$ respectively.  Then, we want to show that the count $\# \mathcal{M}(X, \{g_t\})$ agrees with $\# \mathcal{M}(X,\{k_t\})$.  (Technically, if $f$ is orientation-reversing on $H_2^+(X;\mathbb{R})$, spanned by surfaces with positive self-intersection, then the counts only agree mod 2.)  If we have a path $\gamma$ from $g$ to $k$, then $f^* \gamma$ is a path from $f^*g$ to $f^*k$, and we see that $\# \mathcal{M}(X, \gamma) = \# \mathcal{M}(X, f^*\gamma)$ since they are related by $f$.  There is a loop obtained by concatenating $g_t$, $f^*\gamma$, $k_{1-t}$, $\gamma_{1-t}$.  This bounds a disk and so we see the sum of associated counts is exactly zero.  Since the contributions from $\gamma$ and $f^*\gamma$, we get the desired invariance. 

Even if you didn't care.  Since the invariant did not depend on choice of metric $g$ or path from $g$ to $f^*g$, we will write this moduli count as $I(X,f)$.  This satisfies lots of great properties, such as being an isotopy invariant, $I(X,f' \circ f) = I(X,f') + I(X,f)$, and $I(X,f^{-1}) = -I(X,f)$.  Hence, if $f$ is smoothly isotopic to the identity, then since $id \circ id = id$, we see that $I(X,f) = 0$.  Pretty cool, right? 

So, now we want to use it to distinguish some self-diffeomorphisms.  Consider the manifold $N = \mathbb{C}P^2 \#_2 \overline{\mathbb{C}P^2}$.  There are two embedded spheres of self-intersection $-1$ we are interested in: $S \pm E_1 + E_2$, where $S, E_i$ are copies of $\mathbb{C}P^1$ in each summand.  Here, the sum just means tube together these surfaces in the four-manifold.  There is a "reflection" diffeomorphism associated to each of the resulting spheres.  (I don't actually understand this construction to be honest, but it doesn't matter to understand this post.)  We will compose the two reflection maps, denoted $f_0$, and apply it to $X \# \mathbb{C}P^2 \#_2 \overline{\mathbb{C}P^2}$ for our favorite manifold $X$.  (The composition of the two maps is what allows us to get something orientation preserving on $H^+_2$, so we don't need to worry about signs.)  If $\mathcal{M}(X)$ is 0-dimensional for some bundle over $X$, then there is the AMAZING property that $I(X \# \mathbb{C}P^2 \#_2 \overline{\mathbb{C}P^2}, id \# f_0)$ is $\pm 4$ times $\mathcal{M}(X)$.  (I'm being sloppy with signs because I don't want to talk about homology orientations.)  

There are pairs of homeomorphic $X$ and $X'$ with different Donaldson invariants that become diffeomorphic after connected sum with a single $\mathbb{C}P^2$ (and hence after connected sum with $\mathbb{C}P^2 \#_2 \overline{\mathbb{C}P^2}$).  Furthermore examples can be arranged such that the homology classes of $S, E_1, E_2$ are fixed.  Let $\psi$ be such a diffeomorphism between $X$ and $X'$ after this connected sum.  Then, we can take $$ \phi = \psi^{-1} \circ (id \# f_0)^{-1}_{X'} \circ \psi \circ (id \# f_0)_X$$ to get a self-diffeomorphism which acts by the identity on homology.  Therefore, $\phi$ is topologically isotopic to the identity!  On the other hand, the properties of our diffeomorphism invariants imply that $$I(X \# \mathbb{C}P^2 \#_2 \overline{\mathbb{C}P^2}, \phi) = \pm 4(D(X) - D(X')),$$ where $D$ denotes the 0-dimensional count Donaldson invariant, $\# \mathcal{M}(X)$.  However, if $X$ and $X'$ had different Donaldson invariants (with 0-dimensional moduli spaces),  then we would see that $\phi$ could not be smoothly isotopic to the identity!  These examples do exist, and so we get self-diffeomorphisms which are topologically isotopic but not smoothly isotopic.  In other words, the forgetful map from $\pi_0(Diff) \to \pi_0(Homeo)$ has some kernel for the right four-manifolds.  So cool right?!?!?